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Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case.

square

Now we seek some addition condition on $X\to Y$ that forces the right square is a pullback too.

My question: is epic a sufficient condition? (If the category is Sets, then yes.)

Added: Let $P$ be the pullback of the right square, then there exists $B\to P$, and the square $A\to P \to Y$ // $A\to X \to Y$ is a pullback, so we have the following diagram in which the bottom and the whole squares are pullback, so is the upper square. If the category is Sets, $X\to Y$ is surjective then $A\to P $ is also surjective. Since the pullback of $B\to P$ along a surjective map is an bijection, $B\to P$ must be a bijection. This shows the right square of the original diagram is a pullback. We can also see why we consider some nice condition on $X\to Y$.

2nd square

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    $\begingroup$ I can't tell immediately whether this is research-level. Therefore it probably is. Hmm, but only assuming I actually know anyhing about category theory. The dilemma is unbearable. $\endgroup$
    – Andrej Bauer
    Commented Sep 24, 2013 at 18:19
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    $\begingroup$ @Ma Ming: Can you add your proof for the category of sets when $X \to Y$ is epi? $\endgroup$
    – Martin Brandenburg
    Commented Sep 25, 2013 at 8:22
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    $\begingroup$ In sets every epi splits. I have a feeling it might help to know that $X \to Y$ is a split epi, not just epi. $\endgroup$
    – Andrej Bauer
    Commented Sep 25, 2013 at 8:54
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    $\begingroup$ My initial guess is that your condition is equivalent to the condition that the pullback functor along $X \rightarrow Y$ is conservative. The pullback functor is conservative if the mediating maps are strong epimorphisms (in fact, up to some mild assumptions on the category, the converse is true as well). If my guess is correct, then this explains the case of regular categories --- since in a regular category regular epis coincide with strong epis and are stable under pullbacks. $\endgroup$
    – Michal R. Przybylek
    Commented Sep 26, 2013 at 15:34
  • $\begingroup$ @MaMing, I started writing an answer based on the above comment, but now I see that you've already accepted one of the answers given so far. Does it mean that you are completely satisfied with the answer and not interested in any further elaboration? $\endgroup$
    – Michal R. Przybylek
    Commented Sep 27, 2013 at 21:12

4 Answers 4

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No, epic is not sufficient; there is a counterexample in Pos, the category of posets. Take A, B, and X each to be the discrete two-element poset $\{0,1\}$; take C, Y, and Z to be the same elements with their natural ordering. $\{ 0 < 1 \}$; and take all maps to be identities on underlying sets.

On the other hand, split epic is sufficient in any category; this is a reasonably straightforward diagram chase.

Trying to weaken this, I would suspect (though I haven’t checked it) that in regular categories, regular epic may well be a sufficient condition; certainly regular categories (and strengthenings like exact categories) give a natural setting for these sorts of interactions between epis and pullbacks.

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    $\begingroup$ It seems true. Consider the second diagram of mine. $A\to P$ is a regular epi, and the pullback of $B\to P$ along a regular epi is an iso, we want to show $B\to P$ is also an iso. In fact, let $B\to im \to P$ be the image factorization, whose pullback along a regular epi is regular epi + mono, and their composition is an iso, so each of them should be iso. So the problem reduces to when $B\to P$ is regular epi or mono. It is easy for the case mono using factorization property. (not finished for regular epi) $\endgroup$
    – Ma Ming
    Commented Sep 25, 2013 at 19:29
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    $\begingroup$ Now I get it. By a easy diagram chasing $B\to P$ is a mono, then $A\to B\to P $ is a regular epi, mono factorization of regular epi $A \to P$, so $B\to P$ is a iso. $\endgroup$
    – Ma Ming
    Commented Sep 25, 2013 at 21:03
  • $\begingroup$ As john pointed, if the category is not a regular category, then one may use the fact that $A\to P$ is a regular epi hence a extremal epi to deduce $B\to P$ is an iso. $\endgroup$
    – Ma Ming
    Commented Oct 22, 2013 at 20:10
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I tried to write the explanation of the comment about a dozen of times, but was never satisfied with the result. Finally, I decided to write a full note (I will try to put it on arXiv in a few minutes; here it is) describing the natural setting for such questions:

http://www.mimuw.edu.pl/~mrp/the_other_pullback_lemma.pdf

(the note still needs some improvements, but I am running out of time now...)

I found it easier to characterise your condition by "extremal epimorphisms" rather than "strong epimorphisms" (notice however, that in case of finite connected limits, these concepts coincide). Here is the formal statement:

Let us assume that finite connected limits exist. The following are equivalent:

  • your condition along $e \colon X \rightarrow Y$ holds
  • $e \colon X \rightarrow Y$ is an extremal morphism stable under pullbacks.
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  • $\begingroup$ This is great! Very glad you wrote it and came back later to post this answer $\endgroup$
    – David White
    Commented Nov 11, 2013 at 22:15
  • $\begingroup$ Thanks for your interest! @DavidWhite I write a lot of notes on various questions, but taking into account that I'm a slow writer, most of the time I'm reluctant to post them as answers: thinking that people are not interested in "old questions". You may find interesting another note, that actually grew to a paper: arxiv.org/abs/1306.3703. I'm trying to sell it under a different name, but truly, it is an explanation of my comment on "internal connectives" on question: mathoverflow.net/questions/108397/… $\endgroup$
    – Michal R. Przybylek
    Commented Nov 14, 2013 at 20:32
  • $\begingroup$ Indeed, very interesting. I'm another who likes to dig up old questions and leave long answers (mostly so I can find them in the future). Sometimes it just takes a while for a solution to come to me. I don't see this as a bad thing, and if thinking about it for a longer period of time gives a better answer, then actually I think it's a good thing. Cheers. $\endgroup$
    – David White
    Commented Nov 14, 2013 at 22:22
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Consider the category consisting of that diagram, together with an extra object $P$ with maps $Y\leftarrow P\to C$ that commute with the maps to $Z$. Then in this category, the whole diagram and the left square are pullbacks, $X\to Y$ is epic, but the right square is not a pullback.

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  • $\begingroup$ And how do you compose all the composable arrows? Are you talking about the category which is freely generated by the diagram seen as a directed graph? $\endgroup$
    – Andrej Bauer
    Commented Sep 25, 2013 at 8:53
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    $\begingroup$ I mean the category $\mathcal{C}$ such that a functor out of $\mathcal{C}$ is the same as a commutative diagram of that shape in the target category. So it is generated by the directed graph subject to the relations that say the diagram commutes. $\endgroup$
    – Eric Wofsey
    Commented Sep 25, 2013 at 17:53
  • $\begingroup$ Ah yes, I forgot the diagram implied some relationships. Thanks. $\endgroup$
    – Andrej Bauer
    Commented Sep 25, 2013 at 18:27
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A sufficient condition in a category with pullbacks is that $X \to Y$ be a pullback stable regular epimorphism: a regular epimorphism all of whose pullbacks are also regular epimorphisms. This property holds in any regular category.

Stability implies that $A \to B$ and $A \to P$ are (pullback stable) regular epis too. Thus $B \to P$ is strong epi by 2 out of 3 on the upper square of the lower diagram. So to show $B \to P$ is an isomorphism you just need to show it is mono.

You need to show that the two maps from the kernel pair $B\times_{P}B$ of $B \to P$ coincide. To see this we'll use the map between kernel pairs $A=A \times_{A}A \to B \times_{P} B$ induced by the same square as before. This map is, in fact, the pullback of $A \to P$ along equally the domain or codomain projections for the graph map between kernel pairs, and so a regular epi itself. Thus the two maps $B\times_{P}B \to B$ coincide because their precomposite with this regular epi does.

(There is a good reason for replacing epis in Set by pullback stable regular epis in a category C with pullbacks. The reason is that in any such category C the pullback stable regular epis, taken as singleton covers, determine the basis for a Grothendieck topology on C. Moreover it is subcanonical and, furthermore, the induced functor $C \to Sh(C)$ to the category of sheaves preserves pullbacks and sends pullback stable regular epis to the same. Thus any exactness property which holds between pullbacks and epis in a Grothendieck topos holds between pullbacks and pullback stable regular epis in an arbitrary category with pullbacks.)

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  • $\begingroup$ Wonderful heuristic! But can you say more about why the topology we get is subcanonical? $\endgroup$
    – Zhen Lin
    Commented Sep 26, 2013 at 20:53
  • $\begingroup$ Consider a category with pullbacks and a topology on it in which all covers consists of single arrows. Then all representables are sheaves if and only if all covers are the coequalisers of their kernel pairs. $\endgroup$
    – john
    Commented Sep 26, 2013 at 21:01
  • $\begingroup$ Ah, right. I forgot that regular epimorphisms are effective epimorphisms when kernel pairs exist. Thanks! $\endgroup$
    – Zhen Lin
    Commented Sep 26, 2013 at 21:04

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